An insurance company classifies people as normal or accident prone. Suppose that

Question

An insurance company classifies people as normal or accident prone. Suppose that the probability that a normal person has an accident in a specified year is 0.2 and that for an accident prone person this probability is 0.6. Further suppose that 18% of the policyholders are accident prone. A policyholder had no accidents in a specified year. What is the probability that he or she is accident prone ?

This isn't simple probability. There are some references here,

https://www.coursehero.com/file/pfca3j/In-particular-if-P-A-1-then-P-E-P-A-P-E-A-P-A-C-P-E-A-C-Example-49-At-a/

But i cant access it.

Explanation / Answer

Let

N = normal
P = accident prone
A = has accidents

Hence, by Bayes' Rule,

P(A') = P(N) P(A'|N) + P(P) P(A'|P) = (1-0.18)*(1-0.2) + 0.18*(1-0.6) = 0.728

Hence,

P(P|A') = P(P) P(A'|P)/P(A') = 0.18*(1-0.6)/0.728 = 0.098901099 [ANSWER]

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